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  "body": "\n## Statement\n\nLet $z_1, \\dots, z_n$ be complex numbers that are [[linearly independent|0CRMGPAH83]] over $\\Q$.\nThen the field $\\Q(z_1, \\dots, z_n, e^{z_1}, \\dots, e^{z_n})$ has transcendence degree at least $n$ over $\\Q$.\nHere $e^{z}$ is the [[complex exponential|07GJV36896]].\nThe transcendence degree of a field $K \\supseteq \\Q$ is the largest number of elements of $K$ that are algebraically independent over $\\Q$.\n\n## Remarks\n\nThe conjecture is open.\nThe case $n = 1$ is known: if $z \\ne 0$, then $z$ or $e^z$ is [[transcendental|0VTQXTCR2G]], which is the Hermite–Lindemann theorem.\nIf $z_1, \\dots, z_n$ are algebraic, the conjecture gives the [[Lindemann–Weierstrass theorem|0JNCBFMPXR]], which is proved.\nWith $z_1 = 1$ and $z_2 = i\\pi$, it would imply that $e$ and $\\pi$ are algebraically independent; it is not even known whether $e + \\pi$ is irrational.\nIt would also imply the [[Gelfond–Schneider theorem|RN58Z2NWAC]] and [[Baker's theorem|WMBJSNG57F]], and even that $\\Q$-linearly independent logarithms of algebraic numbers are algebraically independent.\n\n## Explanations\n\n### L1 human\n\nTake numbers $z_1, \\dots, z_n$ with no rational linear relation among them, together with their exponentials.\nThe conjecture says that among these $2n$ numbers, at least $n$ are transcendental and free of any polynomial relation.\nIt is a single statement that would contain most known transcendence results and many famous open ones.\n",
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      "posed_by": "Stephen Schanuel"
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